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In RD Sharma Class 12 Chapter 3 Exercise 3.9, there are seven questions you can refer to from this material. These are Level 1 questions that are explained in a detailed manner and can be beneficial for revision.

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The RD Sharma solution is solely intended to provide students with a comprehensive and detailed explanation of the chapter. The solution has answers to each of the exercise's questions in a logical order. Students can gain a competitive advantage by referring to RD Sharma Class 12 Solutions Chapter 3 Ex 3.9, which will help them get better grades.

What is Inverse Trigonometry?

Trigonometric functions are the values that define relations concerning a triangle. Inverse trigonometry is the inverse of those functions that helps determine the third side if two sides are known. Inverse trigonometry is used in physics, navigation, and programming. For more information, check RD Sharma Class 12 Solutions Inverse Trigonometric Functions Ex 3.9.

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As RD Sharma and NCERT books follow the same CBSE syllabus, you can solve one book if you are perfect in the other. Therefore, once you finish this material, you will be able to solve NCERT problems with ease. Moreover, as RD Sharma Class 12 Solutions Inverse Trigonometric Functions Ex 3.9 contains essential questions, you might find similar questions in the NCERT material.

RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on 21 Jan 2022, 12:37 PM IST

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Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.9

Inverse Trigonometric Function exercise 3.9 question 1 (i)

Answer:
$\frac{24}{25}$
Hint:
Take the negative sign out of the sine inverse bracket and after that cosine of a negative angle will be equal to cosine of the same positive angle.
This way you get rid of the negative sign
Concept:
Inverse Trigonometry
Solution:
$cos\, cos((\frac{-7}{25}))=cos\, cos(-(\frac{7}{25}))$ $(-x)=-(x)$
$=cos\, cos((\frac{7}{25}))$ $[cos\, cos(-\theta )=cos\, \theta ]$
$Let\: \: \: sin^{-1}(\frac{7}{25})=x$.......(1)
$sin\, x=\frac{7}{25}$
$cos\, cos\, x=\sqrt{1-x}$ $[x+x=1]$
$=\sqrt{1-(\frac{7}{25})^{2}}$
$=\sqrt{1-\frac{49}{625}}$
$=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}=\frac{24}{25}$
$\therefore x=cos^{-1}(\frac{24}{25})$.......(2)
From (1) and (2)
$sin^{-1}(\frac{7}{25})=cos^{-1}(\frac{24}{25})$
$\therefore cos(sin^{-1}(\frac{7}{25}))=cos(cos^{-1}(\frac{24}{25}))$
$\therefore cos(sin^{-1}(\frac{7}{25}))=\frac{24}{25}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer trigo ratio using identities.

Inverse Trigonometric Function exercise 3.9 question 1 (ii)

Answer:
$\frac{13}{5}$
Hint:
Convert the cot^-1 term to a tan^-1 term so that we can us the following result to solve the sum
$sec^{2}(x)=1+tan^{2}(x)$
$\therefore sec\, (x)=\sqrt{1+tan^{2}(x)}$ ...using this we can simplify the sum
Concept:
Inverse Trigonometry
Solution:
$sec[cot^{-1}(\frac{-5}{12})]=sec[tan^{-1}(-\frac{12}{5})]$ $[cot^{-1}(x)=tan^{-1}(\frac{1}{x})]$
$=sec[-tan^{-1}(\frac{12}{5})]$ $[tan^{-1}(-x)=-tan^{-1}(x)]$
$=sec[tan^{-1}(\frac{12}{5})]$ $[sec(-\theta )=sec\, \theta ]$
$=\sqrt{1+tan^{2}(tan^{-1}(\frac{12}{5}))}$ $[sec^{2}\theta =1+tan^{2}\theta ]$
$=\sqrt{1+(\frac{12}{5})^{2}}\; \; \; =>\sqrt{1+\frac{144}{25}}\; \; \; =\sqrt{\frac{25+144}{25}}$
$=\sqrt{\frac{169}{25}}$
$=\frac{13}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 1 (iii)

Answer:
$\frac{-5}{12}$
Hint:
Eliminate the negative sign by using formula sec^-1(-x)=$\pi$-sec^-1(x) and then convert the sec^-1 term into a cot^-1 term to simplify the sum
Concept:
Inverse Trigonometric
Solution:
$cot(sec^{-1}(\frac{-13}{5}))=cot(\pi -sec^{-1}(\frac{13}{5}))$ $[sec^{-1}(-x)=\pi -sec^{-1}(x)]$
$=-cot(sec^{-1}(\frac{13}{5}))$ $[cot(\pi -\theta) =-cot\, \theta ]$
Let
$sec^{-1}(\frac{13}{5})=x$ ..............(1)
$sec\, x=\frac{13}{5}$
$\therefore tan\, x=\sqrt{sec^{2}x-1}$
$=\sqrt{(\frac{13}{5})^{2}-1}\: \: \: =\frac{12}{5}$
$cot\, x=\frac{5}{12}\Rightarrow x=cot^{-1}(\frac{5}{12})$ ............(2)
$sec^{-1}(\frac{13}{5})=cot^{-1}(\frac{5}{12})$
From (1) and (2)
$=cot(sec^{-1}(\frac{-13}{5}))$
$=-cot(sec^{-1}(\frac{13}{5}))$
$=-cot(cot(\frac{5}{12}))$
$=\frac{-5}{12}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 2 (i)

Answer:
$\frac{-24}{7}$
Hint:
Get rid of the negative sign by using the formula
$cos^{-1}(-x)=\pi -cos^{-1}(x)$
Convert the cos^-1 term to a sec^-1 term.
Now use the formula
$sec^{2}(x)=1+tan^{2}(x)$
$tan(x)=\sqrt{sec^{2}(x)-1}$
Concept:
Inverse Trigonometry
Solution:
$tan(cos^{-1}(\frac{-7}{25}))=tan(\pi -cos^{-1}(\frac{7}{25}))$ $[cos^{-1}(-x)=\pi -cos^{-1}(x)]$
$=-tan(\pi -cos^{-1}(\frac{7}{25}))$ $[tan(\pi -\theta )=-tan\theta ]$
$=-tan(sec^{-1}(\frac{25}{7}))$ $[cos^{-1}(x)=sec^{-1}(\frac{1}{x})]$
$=-\sqrt{sec^{2}(sec^{-1}(\frac{25}{7}))-1}$ $[tan^{2}\theta =sec^{2}\theta -1]$
$=-\sqrt{\frac{625-49}{49}}$ $=-\frac{24}{7}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 2 (ii)

Answer:
$\frac{13}{5}$
Hint:
Get rid of the negative sign by using the formula
$cos^{-1}(-x)=\pi -cot^{-1}(x)$
Convert the cos^-1term to a sec^-1 term.
Now use the formula
$cos\, ec^{2}(x)=1+cot^{2}(x)$
$cos\, ec(x)=\sqrt{cot^{2}(x)+1}$
Concept:
Inverse Trigonometry
Solution:
Let
$cot^{-1}(\frac{-12}{5})=x$
$cot\, x=\frac{-12}{5}$
$cos\, ec^{2}(x)=1+cot^{2}(x)$
$cos\, ec(x)=\sqrt{cot^{2}(x)+1}$
$=\sqrt{1+(\frac{12}{5})^{2}}$
$=\sqrt{1+(\frac{144}{25})}$
$=\sqrt{\frac{169}{25}}\; \; \; \; \;= \frac{13}{5}$
$cos\, ec(x)= \frac{13}{5}$
$x=cot^{-1} (\frac{-12}{5})$
$cos\, ec(cot^{-1} (\frac{-12}{5}))=\frac{13}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 2 (iii)

Answer:
$\frac{4}{5}$
Hint:
Get rid of the negative sign by using the property
$tan^{-1}(-x)=-tan^{-1}(x)$
Now use this property to simplify
$cos(-x)=cos(x)$
Now find the sec^-1 equivalent of the given tan^-1 term
Now use
$cos(x)=\frac{1}{sec (x)}$$cos(x)=\frac{1}{sec (x)}.\: \: to \; solve\: the\: sum$
Concept:
Inverse Trigonometry
Solution:
Let
$tan^{-1}(\frac{-3}{4})=x$
$tan\, x=\frac{-3}{4}$
$sec^{2}x=1+tan^{2}x$
$sec\, x=\sqrt{1+tan^{2}x}$
$=\sqrt{1+(\frac{-3}{4})^{2}}$
$=\sqrt{1+\frac{9}{16}}\; \; \: =\sqrt{\frac{25}{16}}$
$\therefore sec(x)=\frac{5}{4}$
$\therefore cos(x)=\frac{1}{sec(x)}\; \; \; =\frac{4}{5}$
$But \; \; x=tan^{-1}(\frac{-3}{4})$
$cos(tan^{-1}(\frac{-3}{4}))=\frac{4}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 3

Answer:
$\frac{-56}{65}$
Hint:
Get rid of the negative sign by using the property
$cot^{-1}(-x)=\pi -cot^{-1}(x)$
$cos^{-1}(-x)=\pi -cost^{-1}(x)$
Use formula
$sin(A+B)=sinAcosB+cosAsinB$
Concept:
Inverse Trigonometry
Solution:
Let
$cos^{-1}\frac{3}{5}=\theta _{1}$ ....(1)
$cos \: \theta _{1}=\frac{3}{5}$
$\theta _{1}=sin^{-1}(\frac{4}{5})$ ....(2)

Let
$cot^{-1}(\frac{5}{12})=\theta _{2}$ ....(3)
$cot\, \theta _{2}=\frac{5}{12}$
$\theta _{2}=cos^{-1}(\frac{5}{13})\; \; =sin^{-1}(\frac{12}{13})$ ....(4)

$sin\, sin((\frac{-3}{5})+cot\, cot(\frac{-5}{12}))$
$=sin\, sin(\pi -(\frac{3}{5})+\pi -(\frac{5}{12}))$
$=sin\, (2\pi -(cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12})))$
$=-sin\, (cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12}))$
$=-\left[\sin \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \cos \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)\right]$
$=-\left[\sin \left(\sin ^{-1}\left(\frac{4}{5}\right)\right) \cdot \cos \left(\cos ^{-1}\left(\frac{5}{13}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)\right]$
From (1), (2), (3) and (4)
$=-[\frac{4}{5}\times \frac{5}{13}+\frac{3}{5}\times\frac{12}{13} ]$
$=[\frac{-56}{65} ]$

Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

tion of all trigonometric functions and their simplification.

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As inverse trigonometric functions contain many questions, it is best to divide your work and then do it systematically. Solving all of them at once is not possible because there are many concepts that you will have to remember. RD Sharma Class 12th Exercise 3.9 provided by Career360 is an excellent source for students to save time and learn this chapter as efficiently as possible.

RD Sharma Class 12th Exercise 3.9 solutions will provide you a thorough insight on the themes and the questions. Inverse trigonometry has applications in various domains, including engineering, geometry, and physics, making it a fundamental unit for Class 12 students.

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